What is ((x+2)^2)/(16)+((y-2)^2)/(12)=1 ?

The solution to ((x+2)^2)/(16)+((y-2)^2)/(12)=1 is Ellipse with (h,k)=(-2,2),a=4,b=2sqrt(3)

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((x+2)^2)/(16)+((y-2)^2)/(12)=1

\frac{( x + 2 ) ^{2}}{16} + \frac{( y - 2 ) ^{2}}{12} = 1

(h, k) = (- 2, 2), a = 4, b = 23

Solution steps

\frac{( x + 2 ) ^{2}}{16} + \frac{( y - 2 ) ^{2}}{12} = 1

Rewrite

\frac{( x + 2 ) ^{2}}{16} + \frac{( y - 2 ) ^{2}}{12} = 1

in the form of the standard ellipse equation

\frac{( x - ( - 2 ) ) ^{2}}{4 ^{2}} + \frac{( y - 2 ) ^{2}}{( 2 3 ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 2, 2), a = 4, b = 23

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 2, 2), a = 4, b = 23

2 x^{2} + y^{2} + 10 x + 12 y = - 12

\frac{x ^{2}}{15} + \frac{y ^{2}}{4} = 1

5 x^{2} + y^{2} = 25

4 (x - 5)^{2} + 3 (y - 5)^{2} = 192

25 x^{2} + y^{2} = 1

What is ((x+2)^2)/(16)+((y-2)^2)/(12)=1 ?
The solution to ((x+2)^2)/(16)+((y-2)^2)/(12)=1 is Ellipse with (h,k)=(-2,2),a=4,b=2sqrt(3)